Know the functioning of the function?

Algebra Level 4

Let $f(x)$ be a polynomial. It is known that for all $x$ ,

$\large f(x)f(2x^2) = f(2x^3+x)$

If $f(0)=1$ and $f(2)+f(3)=125$ , find $f(5)$ .

The answer is 676.

1 solution

Ravi Dwivedi
Jul 6, 2015

Let be a root of f(x).

Then by given is also a zero of f.

Triangle inequality implies that

This gives infinitely many zeros by which is a contradiction.

Comparing coefficinets of both sides of original given equation, we conclude that both the leading coefficient and f(0) are 1. f(0)=1=product of all zeros (by vieta's rule) in view of above contradiction and the fact that product of roots is 1, we get that all roots have absolute value=1. So

Then also

We conclude that

From f(2)+f(3)=125 we get n=2

This yields f(5)=676

Note that $f(x) \equiv 0$ is not a solution, because it is given that $f(0) = 1$ .

Calvin Lin Staff - 5 years, 11 months ago

Sorry!! Will take care in future

Ravi Dwivedi - 5 years, 11 months ago

Yes u r right!! So what's the approach without assuming it a polynomial?

Ravi Dwivedi - 5 years, 11 months ago

The times that i've seen this problem (or similar versions of it), I believe that it states the assumption of a polynomial (of finite degree)

In terms of finding functions that satisfy the functional equation, we could do stuff with algebraic closure of $\sqrt{ 2 }$ , and define the function at those values to be $( 1 + x^2)^2$ , while the function everywhere else is 0.

Of course, this doesn't get around that the question is asking for $f(5)$ . So, maybe there is another way to only solve this for $\mathbb{Q} ( \sqrt{2})$ , without resorting to polynomial arguments? (not sure)

Calvin Lin Staff - 5 years, 11 months ago

Can you please explain this in a more shorter and simpler way? Because I get confused at some places, like the step where you used the 'triangle' inequality and then wrote alpha n+1 in terms of alpha n..

Anu Radha - 2 years, 10 months ago

Your proof has several shortcomings.

First, having $|2\alpha^2+1|=1$ and $|\alpha|=1$ effectively implies that $\alpha^2=-1$ but you did not prove it.

Second, you proved that the only possible roots of $f$ are $i$ and $-i$ but you did not prove that they have the same multiplicity and I don't even know whether it's true (currently I'm still working at it).

Maxence Seymat - 1 year, 5 months ago

The first shortcoming was dealt with in showing $1 = |2\alpha^2 + 1 | \geq |2\alpha^2| - 1 = 1$ which implies (skipping a minor step) $\alpha^2 = -1$ .
The second shortcoming is a valid criticism. We can demonstrate that multiplicity needs to be the same via setting $f(x) = ( x - i) ^a ( x+i)^b$ and evaluating when $f(x) f(2x^2) = f(2x^3 + x)$ . After verifying that $(x^2 + 1) \left( (2x^2) ^2 + 1\right) = \left( (2x^3 + x)^2 + 1\right),$ we can divide out by $(x^2+1)$ and so we only need to check when $f(x) = (x-i)$ or $f(x) = (x+i)$ , neither of which hold.

Calvin Lin Staff - 1 year, 5 months ago

Alright for the second shortcoming, I was about to post my solution involving a Taylor expansion (in order to avoid massive coefficients) to the first order when injecting $f(x)=(x-i)^a(x+i)^b$ in $f(x)f(2x^2)=f(2x^3+x)$ , yielding $a-b=0$ .

About the first shortcoming, I wasn't criticising the premises but rather the implication, which isn't that minor to me. You need to remind that $|\alpha|=1$ and then split up the square of $|2\alpha^2+1|$ using complex conjugation, to show that $\mathfrak{Re}(\alpha^2)=-1$ and finally use $|\alpha^2|^2=1$ to get $\mathfrak{Im}(\alpha^2)=0$ , yielding effectively $\alpha^2=-1$ . Please let me know if you use a shorter way to calculate the intersection of two circles in the complex plane.

Maxence Seymat - 1 year, 5 months ago

@Maxence Seymat – For the first shortcoming, thinking about it as the intersection of 2 circles in the complex plane is indeed one way to approach it (and that's the gut reaction that I had since this brute force approach of setting $\alpha = a + bi$ is guaranteed to work).

Do you understand what I wrote up above? Ravi used a different method that happened to work in this case. From (a modified version of) the triangle inequality, we get that

$1 = | 2 \alpha^2 + 1| \geq |2 \alpha^2| - |1| = 2 |\alpha|^2 - 1 = 1$

Hence, equality must hold throughout, hence $\alpha^2$ is in the opposite direction from 1, which means that $\alpha^2 = - 1$ . (That's the minor step that I referenced).

Calvin Lin Staff - 1 year, 5 months ago

@Calvin Lin – I get it now, it's the equality case of the second triangle inequality. But this wasn't obvious to spot at all.

Maxence Seymat - 1 year, 4 months ago

Know the functioning of the function?

The answer is 676.

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